Inverse Problems 6 (1990) 923-934. Printed in the UK
Exact solutions to a linearized inverse boundary value
problem
Margaret Cheneyt , David Isaacsont and Eli L Isaacson:
t Department of Mathematical
Sciences, Rensselaer Polytechnic Institute, Troy, NY
1218a3590, USA
$ Department of Mathematics, University of Wyoming, Laramie, WY 82071, USA
Received 1 June 1990
A b s t r a c t . We consider the problem of determining the spatially varying conductivity of a two-dimensional body from electrical measurements made on the boundary.
A method for solving the linearized problem was suggested by Calder6n. His method
was used previously to solve the linearized problem in the case of two homogeneous
concentric conducting discs with different conductivities. This paper presents the
exact solution to the linearized problem for the case of three concentric discs. As in
the case of two discs, the locations of the discontinuities are recovered exactly.
1. Introduction
In a body B with electrical conductivity e, the electric potential
according to
V . (aVu)= 0
in B .
U
distributes itself
(1.1)
We consider the problem of determining a spatially varying conductivity in a twodimensional body B from measurements made on the boundary S of B.
In [l]Calder6n proposed a method by which one can obtain an approximation to
e if e is close to a constant. More recently it was shown in [3] that Calder6n’s approximate reconstruction yields useful information even when e is not a small perturbation
of a constant, at least in one class of examples. The examples studied consist of twodimensional bodies composed of two concentric discs that determine two regions with
Figure 1
0266-5611/90/060923+12$03.50
Figure 2
@ 1990 IOP Publishing Ltd
923
924
M Cheney, D Isaacson and E L Isaacson
constant conductivities (see figure 1). For these examples the Calder6n reconstruction is composed of infinitely many concentric rings of different conductivities, the
outermost ring having exactly the right radii.
In this paper, we extend the two-disc result to bodies composed of three concentric
discs (figure 2). This class is interesting because it allows us t o study the reconstruction
of an object ‘shielded’ by a ring of different conductivity.
The reconstruction for the three-disc case again is composed of infinitely many
concentric rings, the outermost one having exactly the right radii and conductivity.
However, the inner disc is more difficult to locate, since it is embedded in the sequence of rings generated by the outermost discontinuity. Nevertheless we will see
that the correct locations and conductivity ratios can be obtained from the Calder6n
reconstruction.
2. Review of CalderBn’s method
In this section we establish notation and review the two-dimensional formulation [a, 31
of Calder6n’s method.
We denote by u(p) the electric potential at a point p in B. The potential U is
determined by the current density j through the equations
V . (aVu)= 0
dU
6-
dv
=j
in
B
(2.1)
on S
where U denotes the outward unit normal to S . For equations (2.1), (2.2) t o have a
solution, the current density must satisfy the compatibility condition
c
Also, since the electric potential is defined only up to an additive constant, we choose
the normalization
u(p) d S = 0 .
(2.4)
The Calder6n method for solving the linearized inverse problem is outlined below.
We use the following notation. We write a = a,(l
a), with uo a positive constant.
Also, for any = (Il,E z ) E R 2 , we write (’- = (-t2,El) and z = (< i ( l ) /2. We
define U to be the solution of equations (2.1), (2.2), (2.4) corresponding to the current
density
+
<
j = -iz’ .uexp(-iz.p)
.
We define v(p) = at1exp (-iT.p) so that v satisfies
0 . (aoVv) = 0
and
dv
go--
dv
f
= J
in
B
+
(2.5)
A linearized inverse boundary value problem
925
where
-
j =
8
(exp (-iF.p))
on S.
dv
Here the bar denotes complex conjugation. Then the Calder6n approximation y to
the perturbation 77 is
Y(P) = - ( 2 W 60
J
R2
2 IEI-2
c ( E ; 77) exp (iE . P ) d5
where C(<;11) is found by surface measurements of
C ( E ; 77) =
1
s
U
(2.8)
and v via the formula
( v j - U;) dS.
(2.9)
Equation (2.8) is derived in [2,3] as well as in appendix D. The derivation uses the
approximation
u(p) M uilexp(-iz - p )
which is exact when 77 = 0
3. The three-ring example
+
In this section, we compute C and y if the conductivity U = co(l v) is composed of
three concentric rings. In particular, we consider the conductivity U defined as follows.
For 0 < T~ < r1 < 1 and cl,cz > 0, let
To compute C ( < ; v )and y , we solve equations (2.1), (2.2), (3.1) by separation of
variables. To this end, we introduce polar coordinates p = ( P cos 0 , P sin 0) and then
express the boundary data (2.5) in terms of Fourier series:
(3.4)
(3.5)
926
M Cheney, D Isaacson and E L Isaacson
where the characteristic resistivities pn and p: are given (for n
# 0) by
+
where A+ = 1 ua/ui-l and X i = 1 - ui/ui-l for i = 1 , 2 . Formula (3.7) for pn is
derived in appendix A.
We can now compute C ( [ ;q ) from (2.9) using (3.2), (3.3), (3.4), (3.5), and orthogonality of the exponentids:
The Fourier coefficients c, and En can be computed explicitly (see appendix B):
where ,B is defined by [ = (141 cosp, 141 sin@). Using equations (3.6), (3.9) and (3.10)
in (3.8) gives
(3.11)
The difference in characteristic resistivities appearing in (3.11) can be computed
from (3.6) and (3.7) to be
(3.12)
for n > 0. We divide numerator and denominator of (3.12) by A f A t and use the
notation pi = X;/Xf
= (ui-l - ua)/ (ui-l ui),obtaining
+
(3.13)
Having computed C ( [ ;q ) , we now turn our attention to the Calder6n approximation y. Instead of y itself, we consider its Fourier transform which is given by
(3.14)
927
A linearized inverse boundary value problem
As in [3], we obtain i. as the sum of transforms of characteristic functions. In particular, let xa denote the function that is unity on a disc of radius b centred at the origin
and zero elsewhere. Then its Fourier transform i bis given by [3]
(3.15)
To compute (3.14), we expand (3.13). Provided that
j P1PZ
(pz/r1>2n
- (Ply:"
1
+PZ4") < 1
(3.16)
we can expand (3.13) in a geometric series:
2n
m
(3.17)
where we have written a, = p l r f n
in a binomial series, obtaining
w
+ p z r i n . We next expand the mth term of (3.17)
m
m-k
(3.18)
m=O k=O
Finally we expand ah in (3.18) in a binomial series, yielding
(3.19)
Because an is a sum of two terms, (3.19) splits into the following two terms:
and
Using (3.20) and (3.21) in (3.11) and (3.14), we see that
also is the sum of two terms
i.l and =y2:
(3.22)
(3.23)
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M Cheney, D Isaacson and E L Isaacson
In (3.22) and (3.23) we compare the sum over n to (3.15), which gives us
From (3.24) and (3.25), we conclude that
We interpret this formula in the next section. We recall that (3.26) is derived under
the assumption (3.16). We discuss this assumption in appendix C.
4. Interpretation of results
+
We have just obtained the formula (3.26) for y where a0(l y) is the Calder6n
approximation to the conductivity U . In this section we interpret formula (3.26) and
consider some special cases. Recall that (3.26) is derived under the assumption (3.16).
The formula (3.26) can be written
where
First, the discontinuities in the approximation y to q occur on a sequence of
concentric circles that cluster only at the origin. Between consecutive circles, y is
constant. The outermost circles in the sums I and I1 are r = r1 and T = r 2 , respectively. Consequently, the position of the outer discontinuity r = r1 of r] is determined
exactly, as is q outside the discontinuity. Moreover, the discontinuities in the sum
I that occur outside T = r2 come from precisely the terms with m = k = t for
m = 0 , 1 , 2 , ,[logr2/logrl - I], where [z] denotes the greatest integer less than or
equal to 2. The corresponding circles have radii r r t ' that decrease geometrically.
The inner discontinuity of r] at r = r2 can be determined precisely from the fact that
it breaks this geometric pattern (unless r2 is a power of rl).
A linearized inverse boundary value problem
929
The conductivity ratios ‘ T ~ / ‘and
T ~‘ T ~ / ‘ T ~
also can be determined. In fact, the
value of y in the annulus ry < r < rytl is (for the values of m specified above)
y = - 2 p y t 1 . Thus, pl is determined. Similarly, pz can be determined using the first
term of the sum I I (with m = k = C = 0). The conductivity ratios are obtained from
p1 and pz through the formulae
‘Tz - 1 - Pz
-
1+PZ
‘T1
In [3], the conductivity
‘T
satisfies
‘T
= a,(l
+ 77) where u0 = 1 and
for IPl > r1
for 0 5 Ipl < r1 .
0
V(P) = { a
The approximation y to q is given by
03
m=O
+
where p = - u / ( 2
U ) and x b is the characteristic function of the disc Ipl 5 b. (Note
that we have reversed the sign of p in [3] to be consistent with the notation used in
the present paper .)
The special cases that we consider are:
(1)
771
=0
(‘To
= (71)
(2)
772
=0
(61 = ‘Tz)
(3)
r1 = r2
(4)
r1 = 1
(5)
rz = 0.
All of these are related to the problem of two concentric discs. In cases ( l ) ,( 2 ) above,
the formula for y reduces (as expected) to the formula for two concentric discs:
M
m=O
03
The formulae for the other three cases are:
(3)
(4)
(5)
930
M Cheney, D Isaacson and E L Isaacson
Each of these indicates that the ring structure of the two-disc case is preserved. However, the conductivity ratio does not agree with that recovered from the two-disc
formula.
Acknowledgments
We thank Michael Vogelius for asking about the three-disc case. The work of Margaret
Cheney was supported in part by the Office of Naval Research Young Investigator
grant N00014-8SK-0224 and grant N00014-89-J-1129. The work of David Isaacson was
supported in part by the NSF (BCS-8706340) and the NIH (GM 39388 and GM 42935).
Appendix A
In this appendix we compute the p n , which are the eigenvalues of the map
@ U / W
Ir=l -+ U L l corresponding to the eigenfunctions exp(in0). Here the boundary map is restricted to functions satisfying (2.3) and (2.4).
We wish to solve equations (2.1), (3.1) with boundary data
for n 2 1. (For n = 0, the right side of (A.l) does not satisfy condition (2.3); for
n 5 -1, we can replace 0 by -0.) Using separation of variables, we write the potential
U as u(r,0) = U(r)exp(inB), and we note that U(1) is the desired eigenvalue.
The function U has the form
+
AOrn
rl<rll
r2 < r 5 rl
0 5 r 5 rz.
Here the coefficients A i , B, are determined by the boundary condition
a
;
'
= U'(1) = n ( A , - B o )
(A.3)
and the matching conditions at r1 and r2:
These conditions represent continuity of
be written compactly as
U
and u ( a u / a u ) . Equations (A.4)-(A.7) can
A linearized inverse boundary value problem
93 1
where M = MI M , with
(A.9)
and Ai = Q ~ / Q ~for
- ~i = 1 , 2 .
To find U ( l ) , we need only compute
U(1) = A, + B o = ( 1 , l ) M
(A.lO)
In this notation, the boundary condition (A.l) is
= A , - Bo = A,(l, - l ) M
(3
.
(A.ll)
We solve ( A . l l ) for A, and substitute the result into (A.10), obtaining
where A t = 1+ A i and A T = 1 - Xi.
Appendix B
In this appendix we verify formulae (3.9) and (3.10). We do this in several stages.
First, we derive (3.10) from the Fourier series (3.5). Then formula (3.3) follows because
v ( r ,0) = exp (-iZ.p). Formula (3.2) differs from (3.3) only in that c,(()
depends
upon z in the same way that E,([) depends upon 7. We content ourselves with
deriving (3.10).
Let
+CO
~ , v ( l , 0 )= exp(-iF.p) =
d,exp(in0)
n=-m
where
JzTexp (-iT.p)exp(-inO)
Here,
d0
n#O
n = 0.
932
M Cheney, D Isaacson and E L Isaacson
Then
IEl i(p-e)
r e p = -e
2
Using this in the expression for d,, we obtain for n # 0
d, = 2a / T-= exp ( - y e i @ - ' ) )
exp(-in6') do.
Next, we expand the first exponential and interchange the order of summation and
integration, obtaining
00
d, =
m=O
& S_:
1
(- y e i @ )
m!
exp(-imO) exp(-inO) d6'
In this expression we use the orthogonality of the exponentials to obtain
Appendix C
Expansion (3.26) holds under the condition (3.16), which we rewrite as
(PI
+ 4,)
+ P1P2T2n( < 1
(C.1)
where T = r 2 / r 1 . This condition holds in two simply stated cases: (i) r2 sufficiently
small; (ii) p 1 and p2 both positive.
Case (i). Because Ipll,
side of (C.l) by
1 ~ ~ r1
1 , and
T
are always less than 1, we can estimate the left
when
Case (ii).
This case occurs if u2 < u1 < uo. First, note that
PlP2
because 0
- w l n-
> -1
(C.4)
< r1 < 1 implies
0 < (1 - pl.;") (1 - P2.2"")
< 1 - Pl$
- P2?
+ P1P2 ( T Z / T l )
2n
.
Second,
PIP2 ( r 2 / r J 2 , < 1 < 1
+
+
P2$.
Condition (C.l) now follows from (C.4) and (C.5).
(C.5)
933
A linearized inverse boundary value problem
Appendix D
In this appendix we derive equation (2.8), which gives the Calder6n approximation y
to q. First, we obtain an alternate expression for C, defined in (2.9), which involves
the Fourier transform f j of q.
From (2.1), (2.6) and Green's Theorem, we obtain
0=
=
J, {vV - ( a V u )- uV. (aoVv)} dp
{v j - u j }
dS -
J, aoqVu- Vu dp.
Thus, C defined in (2.9) can be written
e([; q ) G J, { v j - U ? }
dS =
J, aoqVu Vu dp.
On the right-hand side of ( D . l ) we can approximate
U
by
- p)
u ( p ) x a;' exp(-it
where the error is order 77 when q is small
[a]. Then we have the approximation
Therefore, the identity
VJV(=
~ )u , ' ( - i ~ ) e x p ( - i ~ . p )
when used with (D.2) in the right-hand side of (D.l), yields
J, aoqVu Vu dp M
-
quil(-z
-e1
2uo
- 2 ) exp (-i(z + 5)- p ) dp
qexp(-i[-p)
dp.
B
Here, the error is order q2 when q is small.
If we extend q as 0 outside B, equations (D.l) and (D.3) yield an approximation
to the Fourier transform f j of q:
fj(F)
J
Ra
q(p)exp(-iE ' P ) dP
934
M Cheney, D Isaacson and E L Isaacson
The formula (2.8) for the Calder6n approximation y to q is obtained by taking the
inverse Fourier transform of (D.4):
References
A P 1980 On an inverse boundary value problem Seminar on Numerical Analysis and
its Applications ed W H Meyer and M A Raupp, (Brazilian Mathematics Society: Rio de
Janeiro)
[2] Isaacson D and Cheney M 1990 Effects of measurement precision and finite numbers of electrodes
on linear impedance imaging algorithms Preprint Rensselaer Polytechnic Institute
[3] Isaacson D and Isaacson E L 1989 Comment on Caldenjn’s paper: ‘On an inverse boundary
value problem’ Math. Comput. 52 553-9
[l] Calder6n